# 8.5: Pressure Dependence of Gibbs Energy

The pressure and temperature dependence of $$G$$ is also easy to describe. The best starting place is the definition of $$G$$.

$G = U + pV -TS \label{eq1}$

Taking the total differential of $$G$$ yields

$dG = dU + pdV – pdV + Vdp – TdS – SdT$

The differential can be simplified by substituting the combined first and second law statement for $$dU$$ (consider a reversible process and $$pV$$ work only).

$dG = \cancel{TdS} \bcancel{– pdV} + \bcancel{pdV} + Vdp – \cancel{TdS} – SdT$

Canceling the $$TdS$$ and $$pdV$$ terms leaves

$dG = V\,dp – S\,dT \label{Total1}$

This suggests that the natural variables of $$G$$ are $$p$$ and $$T$$. So the total differential $$dG$$ can also be expressed

$dG = \left( \dfrac{\partial G}{\partial p} \right)_T dp + \left( \dfrac{\partial G}{\partial T} \right)_p dT \label{Total2}$

And by inspection of Equations \ref{Total1} and \ref{Total2}, it is clear that

$\left( \dfrac{\partial G}{\partial p} \right)_T = V$

and

$\left( \dfrac{\partial G}{\partial T} \right)_p = -S$

It is also clear that the Maxwell relation on $$G$$ is given by

$\left( \dfrac{\partial V}{\partial T} \right)_p = \left( \dfrac{\partial S}{\partial p} \right)_T$

which is an extraordinarily useful relationship, since one of the terms is expressible entirely in terms of measurable quantities!

$\left( \dfrac{\partial V}{\partial T} \right)_p = V\alpha$

The pressure dependence of $$G$$ is given by the pressure derivative at constant temperature

$\left( \dfrac{\partial G}{\partial p} \right)_T = V \label{Max2}$

which is simply the molar volume. For a fairly incompressible substance (such as a liquid or a solid) the molar volume will be essentially constant over a modest pressure range.

Example $$\PageIndex{1}$$: Gold under Pressure

The density of gold is 19.32 g/cm3. Calculate $$\Delta G$$ for a 1.00 g sample of gold when the pressure on it is increased from 1.00 atm to 2.00 atm.

Solution:

The change in the Gibbs function due to an isothermal change in pressure can be expressed as

$\Delta G =\int_{p_1}^{p_2} \left( \dfrac{\partial G}{\partial p} \right)_T dp$

And since substituting Equation \ref{Max2}, results in

$\Delta G =\int_{p_1}^{p_2} V dp$

Assuming that the molar volume is independent or pressure over the stated pressure range, $$\Delta G$$ becomes

$\Delta G = V(p_2-p_1)$

So, the molar change in the Gibbs function can be calculated by substituting the relevant values.

\begin{align} \Delta G & = \left( \dfrac{197.0\, g}{mol} \times \dfrac{1\,}{19.32\,g} \times \dfrac{1\,L}{1000\,cm^3} \right) (2.00 \,atm -1.00 \,atm) \underbrace{ \left(\dfrac{8.315 \,J}{0.08206\, atm\,L}\right)}_{\text{conversion unit}}\\ &= 1.033\,J \end{align}

Contributors

• Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)